Question

I have a question related to operator overloading, and it is easy to define a class as well as its operator overloading function as the following codes illustrate:

typedef std::vector<std::vector<int> > ARRAY; 


class ABC
{
public:
    ABC():a(0)
    {
    };
    int a;
    ABC& operator = (int value)
    {
        a = value;
        return *this;
    }
    ABC(int value)
    {
        a = value;

    }
};


void obtain_priority_array(const std::vector<double> &weighting, const ABC &priority_array=NULL)
{

}

int main()
{
    vector<double> weighting;
    weighting.push_back(0.8);
    weighting.push_back(0.9);
    weighting.push_back(0.6);
    weighting.push_back(0.3);
    weighting.push_back(0.5);

    ABC test;
    obtain_priority_array(weighting, test);

    return 0;
}

In the above example, class ABC redefined operator = so that the function void obtain_priority_array(const std::vector<double> &weighting, const ABC &priority_array=NULL) can have a default argument const ABC &priority_array=NULL. My question is if the last parameter in the function comes from STL, for example, const std::vector<int> &priority_array=NULL, how can we redefine operator =. Thanks!

EDIT:
void obtain_priority_array(const std::vector &weighting, const std::vector<int> &sample=NULL) failed!

Answer 1

Your misconceptions start with the proposal to add operator= to allow for a default argument of that type. In your example, it is not operator= being called, but ABC(int).

The reason your code isn’t being accepted when using std::vector is that NULL translates to 0 (at least it does almost all of the time you’ll see it), and the only constructor of std::vector that can take 0, the one taking a count of how many items, is marked explicit.

To fix the immediate problem, the syntax could be changed to:

const std::vector<int> &priority_array = std::vector<int>(0)

However, this introduces different semantics. By your use of NULL, it looks like you were expecting it to represent no vector. This version will provide an empty vector for use if none is given. It will not be no vector at all. If you want that distinction, you should use boost’s optional library, or a simple pointer, as references are not the right tool.